Ncert class 8 maths chapter 2

Linear equation in one variable 

Introduction: Variable : A symbol which can takes various numerical values is called variable.

Ex : x,y…

Term : A Constant alone or A variable alone or combination of constant and variable with multiplication ‘×’ , division  ‘÷’ is called term.

Multiplication ‘×’ ,division  ‘÷’ do not separate the terms.

Ex: x/2y, xy  are single terms.

Expression : An expression is a single term or a combination of terms connected by the symbols ‘+’(plus) or ‘-‘ (minus).

Ex : x²+y is an expression having two terms.

x+y²+z is an expression having 3 terms.

Algebraic Equation : An algebraic equation is equality of algebraic expression involving variables and constants. It has equality sign. Left side of the equality sign is called LHS. Right side of equality is called RHS.

In an equation the values of LHS and RHS are equal. This happens to be true only for certain value of the variable. This is called the solution of the equation.

 x+7=8 is an equation. Solution is x=1.

If degree of equation is ‘1’ then it is called linear equation.

Ex: x+5=6

x²+4x+5=0 is not a linear equation because its degree is 2

x⁴+7x=9 is not a linear equation because it’s degree is 4.

Transposition : In a equation any constant ‘or’ variable ‘or’ term may be taken to the other side with a change in its sign ‘or’ position. This process is called transposition. 

How to solve an equation:

1)We can add the same number on both sides of the equation.

2)We can subtract the same number from both sides of the equation.

3)We can multiply both sides of the equation with same number.

4)We can divide both sides of the equation by the same non-zero number.

When we transpose terms:

‘+’ term becomes ‘ – ‘ term.

‘ – ‘ term becomes’+’term.

Multiplied quantity  becomes divided quantity.

Divided quantity becomes multiplied quantity.

P) x-2=7 then find ‘x’ value.

Ans: From what number ‘2’ should be subtract to get ‘7’.

i.e. x=9        ‘or’

Add 2 on both sides. Then we get x=9.

P) Solve 7x-9=16.

solution: Given 7x-9=16. 

Add ‘9’  on both sides.7x-9+9=16+9

7x=25 Divide both sides by ‘7’

x= 25/7.

Solving verbal (word) problems:

1. Read the problem. Identify the quantity asked for.

2. Select a letter ‘x’ or ‘y’ or ‘z’ etc., to represent the quantity asked for.

3. Change the word statements into symbolic statements.

4. write an equation in the letter selected, showing the relationship in the problem.

5. Solve the equation.

6. Check the answer to make sure that it satisfies the given condition in the problem.

Exercise 2.1

Solve the following equations

1. x-2=7

Given x-2=7

Adding ‘2’ on both sides.

x-2+2=7+2

x=9

2. y+3= 10

Given y+3=10

Subtract ‘3’ on both sides.

y+3-3=10-3

y=7

3. 6 = z+2

Given 6=z+2

Subtract ‘2’ on both sides.

6-2=z+2-2

4=z

$4. \frac{3}{7}+x=\frac{17}{7}$

$\frac{3}{7}+x=\frac{17}{7}$

Subtract ‘3/7’ on both sides 

$\frac{3}{7}+x–\frac{3}{7}=\frac{17}{7}–\frac{3}{7}$

$x=\frac{17–3}{7}=\frac{14}{7}=2$

5. 6x=12

Given 6x=12

Dividing both sides by ‘6’

$\frac{6x}{6}=\frac{12}{6}$

x=2

$6. \frac{t}{5}=10$

Multiplying both sides by ‘5’

$\frac{t}{5}\times 5=10\times 5$

t=50.

$7. \frac{2x}{3}=18$

$Given \frac{2x}{3}=18$

Multiplying both sides by 3/2

$\frac{2x}{3}\times \frac{3}{2}=18\times \frac{3}{2}$

x=27.

$8. 1.6=\frac{y}{1.5}$

Given 1.6=y/1.5

Multiplying  both sides by 1.5

$1.6\times 1.5=\frac{y}{1.5}\times 1.5$

2.4=y

9. 7x-9=16

Given 7x-9=16

Add ‘9’ on both sides.

7x-9+9=16+9

7x=25

Dividing both sides by ‘7’

$x=\frac{25}{7}$

10. 14y-8=13

Given 14y-8=13

Transposing 8 to R.H.S

14y=13+8

14y=21

Dividing both sides by ’14’

$y=\frac{21}{14}$

$y=\frac{3}{2}$

11. 17+6p=9

Given 17+6p=9

Transposing 17 to R.H.S

6p=9-17

6p=-8

Dividing both sides by ‘6’

$\frac{6p}{6}=–\frac{8}{6}$

p=-4/3

$12. \frac{x}{3}+1=\frac{7}{15}$

$Given \frac{x}{3}+1=\frac{7}{15}$

Transposing ‘1’ to R.H.S

$\frac{x}{3}=\frac{7}{15}–1$

$\frac{x}{3}=\frac{7–15}{15}$

$\frac{x}{3}=–\frac{8}{15}$

Multiplying  both sides by 3

$\frac{x}{3}\times 3=–\frac{8}{15}\times 3$

$x=–\frac{8}{5}$

ncert solutions for class 8 maths chapter 2 exercise 2.2

Exercise 2.2

1. If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. what is the number?

solution: Let the required number be ‘x’

According to problem, 

$(x–\frac{1}{2})\frac{1}{2}=\frac{1}{8}$

$x–\frac{1}{2}=\frac{2}{8}$

$x–\frac{1}{2}=\frac{1}{4}$

$x=\frac{1}{4}+\frac{1}{2}$

$x=\frac{1+2}{4}=\frac{3}{4}$

required number is 3/4.

2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

solution: Let breadth = b = ‘x’ m.

Then according to problem length = l =(2x+2) m.

Perimeter of pool = 2 (l+b) =154 m.

2(2x+2+x)=154

2(3x+2)=154

6x+4=154

6x=154-4=150

$x=\frac{150}{6}=25$

length of the pool = 2x+2 = 2(25)+2=52 m.

breadth = x=25 m.

3. The base of an isosceles triangle is 4/3 cm. The perimeter of the triangle is 4 ²/₁₅ cm. What is the length of either of the remaining equal sides?

Solution : let the length of equal sides be x cm.

Given base = 4/3.

Perimeter = x + x + base = 4 ²/₁₅  cm.

$2x+\frac{4}{3}=\frac{62}{15}$

On transposing 4/3 to RHS we obtain.

$2x=\frac{62}{15}–\frac{4}{3}$ $2x=\frac{42}{15}$

On dividing both sides by 2, we obtain

$\frac{2x}{2}=\frac{42}{15}\times \frac{1}{2}$ $x=\frac{7}{5}=1\frac{2}{5}$

∴  The length of equal sides is 1²/₅ cm.

4. Sum of two numbers is 95. If one exceeds the Other by 15, find the numbers.

Solution : Let one number be ‘x’ , 

∴ The other number will be x+15

 According to the question,

x+x+15=95

2x+15=95

2x=95-15

2x=80

x=80/2=40.

The required numbers are x=40 , x+15=55.

5. Two numbers are in the ratio 5:3. If they differ by 18. what are the numbers?

Ans : 

Let the numbers are 5x, 3x respectively.

Given difference between these numbers =18.

5x-3x=18

2x=18

x=18/2=9

1st number= 5x= 5×9=45

2nd number= 3x=3×9=27.

6. Three consecutive integers add up to 51. What are these integers?

Ans: Let three consecutive integers are x,x+1 and x+2.

Sum of these numbers = x + x + 1 + x + 2= 51

3x+3=51

3x=51-3=48

On dividing both sides by 3, we obtain

$\frac{3x}{3}=\frac{48}{3}$

x=16

Hence consecutive integers are 16, 16+1=17 and 16+2=18.

7 The sum of three consecutive multiples of 8 is 888 find the multiples.

solution : Let the consecutive multiples of  8 are 8x, 8(x+1), 8(x+2)

sum of these numbers=8x+8(x+1)+8(x+2)=888.

8x+8x+8+8x+16=888.

24x+24=888

8(3x+3)=888

On dividing both sides by 8

$\frac{8(3x+3)}{8}=\frac{888}{8}$

3x+3=111

On Transposing 3 to R.H.S

3x=111-3

3x=108

On dividing both sides by 3.

$x=\frac{108}{3}$

x=36.

First multiple= 8x=8(36)=288,

Second multiple=8(x+1)=8(36+1)=296,

Third multiple=8(x+2)=8(36+2)=304.

8 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

solution: Let the consecutive integers are x, x+1,x+2

According to problem 2x+3(x+1)+4(x+2)=74

2x+3x+3+4x+8=74

9x+11=74

On transposing 11 to R.H.S

9x=74-11

9x=63

On dividing  both sides by 9

$\frac{9x}{9}=\frac{63}{9}$

x=7

x+1=7+1=8

x+2=7+2=9

The numbers are 7,8 and 9

9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years what are their present ages?

Solution: Let the ages of rahul and haroon are 5x,7x.

Four years later their ages will be 5x+4, 7x+4

According to problem

5x+4+7x+4=56

12x+8=56.

Transposing 8 to R.H.S 

12x=56-8=48

x=3.

10 The number of boys and girls in a class are in the ratio 7:5 the number of boys is 8 more than the number of girls. What is the total class strength?

Solution: Let the number of boys=7x.

Then number of girls =5x

According to problem 

7x=5x+8

Transposing 5x to L.H.S

7x-5x=8

2x=8

x=4

Number of boys=7x=7×4=28

Number of girls = 5x=5×4=20

Total class strength=28+20=48 students.

11 Baichung’s father is 26 years younger than baichung’s grandfather and 29 years older than baichung. The sum of the ages of all three is 135 years. What is the age of each one of them?

Solution : Let Baichung’s father’s age is ‘x’ years.

Then Baichung’s age and baichung’s grand father’s age will be (x-29) years and (x+26) years respectively.

According to problem sum of the ages of three people is 135 years.

x+x-29+x+26=135

3x-3=135

On transposing 3 to R.H.S

3x=135+3

3x=138

On dividing both sides by 3.

$\frac{3x}{3}=\frac{138}{3}$

x=46

Baichung’s father’s age =x  years= 46 years.

Baichung’s age=(x-29)years=(46-29) =17 years.

Baichung’s grand father’s age=(x+26)=(46+26)=72 years

12 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Solution: Let Ravis present age be ‘x’ years.

Fifteen years later, Ravi’s age =4×His present age.

x+15=4x

On transposing ‘x’ to R.H.S

15=4x-x

15=3x

On dividing both sides by 3

$\frac{15}{3}=\frac{3x}{3}$

5=x

Ravi’s present age = 5 years.

13 A rational number is such that when you multiply it by 5/2 and add 2/3 to the product, you get – 7/12 what is the number?

Solution: Let the number be x.

According to problem,

$\frac{5}{2}x+\frac{2}{3}=–\frac{7}{12}$

On transposing 2/3 to R.H.S

$\frac{5}{2}x=–\frac{7}{12}–\frac{2}{3}$ $\frac{5}{2}x=–\frac{15}{12}$

on multiplying both sides by 2/5

x=-15/12×2/5=-1/2

The rational number is -1/2

14  Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100 Rs 50 and Rs 10, respectively. The  ratio of the number of these notes is 2:3:5.  The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Solution: Let the common ratio between the numbers of notes of different denominations be ‘x’.

Numbers of Rs 100 notes, Rs 50 notes and Rs 10 notes will be 2x 3x and 5x respectively.

Amount of Rs 100 notes=Rs(100*2x)=200x

Amount of Rs 50 notes =Rs (50 x 3 x)=Rs150x

Amount of Rs 10 notes= Rs (10 x 5x )=Rs50x

It is given the total amount is Rs 400000.

200x+150x+50x=400000.

400x=400000.

On dividing both sides by 400

x=1000. 

Number of Rs 100 notes=2x=2×1000=2000

Number of Rs 50 notes =3x=3×1000=3000

Number of Rs 10 notes=5x=5×1000=5000

15 I have a total of Rs 300 in coins of denomination Re 1, Rs 2 and Rs 5. The number of Rs 2 coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Solution: Let the number of Rs 5 coins be X.

Number of Rs 2 coins= 3 ×number of Rs 5 coins 3x

Number of Re 1 coins =160 – (Number of coins of Rs 5 and of Rs 2)

=160-(3x+x)=160-4x

Amount of Re 1 coins = Rs[1*(160-4x)]=Rs(160-4x)

Amount of Rs 2 coins=Rs (2×3x)=Rs 6x 

Amount of Rs 5 coins=Rs(5x)=Rs 5x

It is given that the total amount is Rs 300. 

160-4x+6x+5x=300

160+7x=300

On transposing 160 to R.H.S,

7x=300-160

7x=140

on dividing both sides by 7

x=20

Number of Re1 coins=160-4x=160-4×20=160-80=80

Number of Rs 2 coins=3x=3×20=60

Number of Rs 5 coins=x=20.

16 The organisers of an essay competition decide the a winner in the competition gets a prize of Rs 100 and a participant who does not win get a price of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63. 

Solution: Let the number of winner be X.

Therefore, the number of participants who did not pin will be 63 – X.

Amount given to the winner=Rs 100x

Amount given to the participant who did not win =Rs[25(63-x)]=Rs(1575-25x)

According to the given problem

On transposing 1575 to R.H.S

75x=3000-1575

75x=1425

on dividing both sides by 75

$\frac{75x}{75}=\frac{1425}{75}$

x=19 

Number of winners=19

Exercise 2.3

Solve the following equations and check your results.

1 3x=2x+18

Answer: Given 3x=2x+18

on transposing 2x to L.H.S

3x-2x=18

x=18

L.H.S=3x=3(18)=54

R.H.S=2x+18=2(18)+18=36+18=54

L.H.S=R.H.S

The result obtained above is correct.

2 5t-3=3t-5

Answer: Transposing 3t to L.H.S and -3 to R.H.S

5t-3t=-5+3

2t=-2

Dividing on both sides by 2

2t/2=-2/2

t=-1

L.H.S=5t-3=5(-1)-3=-5-3=-8

R.H.S=3t-5=3(-1)-5=-3-5=-8

L.H.S=R.H.S

The result obtained above is correct

3 5x+9=5+3x

Answer: Given 5x+9=5+3x

Transposing 3x to L.H.S and 9 to R.H.S

5x-3x=5-9

2x=-4

Dividing on both sides by 2

$\frac{2x}{2}=\frac{–4}{2}$

x=-2

4 4z+3=6+2z

Answer: Given 4z+3=6+2z

Transposing 2z to L.H.S and 3 to R.H.S

4z-2z=6-3

2z=3

On dividing both sides by 2

$z=\frac{3}{2}$

L.H.S=4z+3=4(3/2)+3=2(3)+3=6+3=9

R.H.S=6+2z=6+2(3/2)=6+3=9

L.H.S=R.H.S

The result obtained above is correct.

5 2x-1=14-x

Answer: Given 2x-1=14-x

Transposing -x to L.H.S and -1 to R.H.S

2x+x=14+1

3x=15

Dividing on both sides by 3

$\frac{3x}{3}=\frac{15}{3}$

x=5

L.H.S=2x-1=2(5)-1=10-1=9

R.H.S=14-x=14-5=9

L.H.S=R.H.S

The result obtained above is correct.

6 8x+4=3(x-1)+7

Answer: Given 8x+4=3(x-1)+7

8x+4=3x-3+7

8x+4=3x+4

On transposing 3x to L.H.S and 4 to R.H.S

8x-3x=4-4

4x=0

x=0

L.H.S=8x+4=8(0)+4=4

R.H.S=3(x-1)+7=3(0-1)+7=-3+7=4

L.H.S=R.H.S

The result obtained above is correct.

$7. x=\frac{4}{5}(x+10)$

Answer: Given

$x=\frac{4}{5}(x+10)$

On multiplying both sides by 5

5x=4(x+10)

5x= 4x+40

Transposing 4x to L.H.S

5x-4x=40

x=40

L.H.S=x=40

$R.H.S=\frac{4}{5}(x+10)=\frac{4}{5}(40+10)=\frac{4}{5}\times 50=4\times 10=40$

L.H.S=R.H.S

The result obtained above is correct.

$8. \frac{2x}{3}+1=\frac{7x}{15}+3$

Answer: Given

$\frac{2x}{3}+1=\frac{7x}{15}+3$

On transposing 7x/15 to L.H.S and 1 to R.H.S.

$\frac{2x}{3}–\frac{7x}{15}=3–1$ $\frac{10x–7x}{15}=2$

$$\begin{gathered} \frac{3x}{15}=2 \\ \frac{x}{5}=2 \end{gathered}$$

Multiplying on both sides with 5

x=10

$L.H.S.=\frac{2x}{3}+1=\frac{2\left(10\right)}{3}+1=\frac{20}{3}+1=\frac{23}{3}$ $R.H.S=\frac{7x}{15}+3=\frac{7\left(10\right)}{15}+3=\frac{14}{3}+3=\frac{14+9}{3}=\frac{23}{3}$

L.H.S=R.H.S

The result obtained above is correct.

$9 2y+\frac{5}{3}=\frac{26}{3}–y$

Transposing y to L.H.S and 5/3 toR.H.S

$2y+y=\frac{26}{3}–\frac{5}{3}$

$3y=\frac{26–5}{3}$

$3y=\frac{21}{3}$

3y=7

dividing both sides by 3

$y=\frac{7}{3}$

$L.H.S.=2y+\frac{5}{3}=2\left(\frac{7}{3}\right)+\frac{5}{3}=\frac{19}{3}$

$R.H.S.=\frac{26}{3}–y=\frac{26}{3}–\frac{7}{3}=\frac{26–7}{3}=\frac{19}{3}$

L.H.S.=R.H.S.

Hence, the result obtained above is correct.

$10 3m=5m–\frac{8}{5}$

$Given 3m=5m–\frac{8}{5}$

Transposing 5m to L.H.S,

$3m–5m=–\frac{8}{5}$

$–2m=–\frac{8}{5}$

Dividing both sides by -2,

$m=\frac{4}{5}$

$L.H.S=3m=3\left(\frac{4}{5}\right)=\frac{12}{5}$

$R.H.S=5m–\frac{8}{5}=5\left(\frac{4}{5}\right)–\frac{8}{5}=\frac{20–8}{5}=\frac{12}{5}$

L.H.S.=R.H.S.

Hence, the result obtained above is correct.

 

Exercise:2.4

1 Amina thinks  of a number and subtract 5|2 from it. She multiplies results by 8. The result now obtained if 3 times the same number she thought of. What is the number?

Solution: Let the number be x.

According to the given problem,

$8(x–\frac{5}{2})=3x$

8x-20=3x

 Transposing 3x to L.H.S. and -22 to R.H.S

8x-3x=20

5x=20

Dividing both sides by 5. 

x=4

Hence, the number is 4

2   A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution: Let the numbers be x and 5x. According to the problem.

5x+21=2(x+21)

⇒5x+21=2x+42

Transposing to 2x to L.H.S. and 21 to R.H.S

5x-2x=42-21

3x=21

Dividing both sides by 3 

x=7

Hence the number are x=7 and 5x=35.

3  Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. what is the two digit number?

Solution: Let x be tens place digit. Then units place digit =9-x.

Then original number=10x+(9-x)1=9x+9

After interchanging the digits

New number=10(9-x)+(x)1=90-9x

According to the problem,

New number=original number+27

90-9x=9x+9+27

90-9x=36+9x

Transposing 9X to R.H.S. and 36 to L.H.S.

90-36=9x+9x

54=18x

Dividing both sides by 18,

3=x and 9-x=9-3=6

Tens place digit =6, units place digit =3

The two digit number =9x+9=9(3)+9=36.

4  One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Solution: Let the digits at tens place and ones place be x and 3x respectively.

Then original number = 10x + 3x=13x

On interchanging the digits, the digits at ones place and tens place will be x and 3x respectively.

Number after interchanging =10×3x+x=30x+x=31x

According to the given question,

Original number+new number=l 88

13 x + 31x=88

44x=88

Dividing both sides by 44,

 x=2 

our original number=13x=13×2=26

 By considering the tens place and ones place as 3x and x respectively,

 The two digit number is obtained is 62.

Therefore the two digit number may be 26 or 62.

 

5  shobo’s mothers present age is 6 times shobo’s present age. Shobo’s age 5 years from now will be one third of his mother’s present age what are their present ages?

Solution: Let shobo’s age be x years.

 Then his mother’s age will be 6x years.

According to the problem,

$After 5yrs shobo‘s age=\frac{shobo‘s mother present age}{3}$

$x+5=\frac{6x}{3}$

x+5=2x

Transposing x to R.H.S,

5=2x-x

5=x

6x=6(5)=30.

∴ The present ages of shobo and shobo’s mother will be 5 years and 30 years respectively.

6  There is a narrow rectangular plot, reserved for school, in Mohuli village. The length and breadth of the plot are in the ratio 11:4. At the rate of Rs 100 per metre it will cost the village panchayat at Rs 75000 to fence the plot. What are the dimensions of the plot?

Solution: 

Given length bredth ratio is 11:4

Let length=11x, breadth=4x.

Perimeter of the plot =2(11x+4x)=2(15x)=30x m

Given that the cost of fencing the plot at the rate of Rs 100 per metre is Rs 75000

∴ 100×perimeter=75000

100×30x=75000

3000x=75000

x=25

Length of the plot=11x=11(25)=275 m.

Breadth=4x=4(25)=100 m.

7  Hasan buys two kinds of cloth material for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that cost him Rs 90 per metre. For every 2 meters of the trouser material he buys 3 metres of the shirt material. He sells the materials at 12 % and 10% profit respectively. His total sale is Rs 36,660. How much trouser material did you buy?

Solution: Given for every 2m of trouser material he buys 3m of shirt material.

Let 2x m trouser material and 3x m shirt material be bought by him.

Per metre selling price of trouser material=

$(90+90\times \frac{12}{100})=Rs. 100.80$

Per metre selling price of shirt material=

$(50+50\times \frac{10}{100})=Rs. 55$

Given that total amount of selling=36,660.

(100.8)×(2x)+55×(3x)=36660.

201.60x+165x=36660

366.60x=36660

Dividing both sides by 366.60

x=100

Trouser material=2x =2(100)=200m.

 

8  Half of a herd of deer are grazing in the field and three fourth of remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Solution: Let the number of deer =x.

Number of deer grazing in the field=x/2

Number of deer playing nearby=3/4×Number of remaining deer.

$=\frac{3}{4}\times [x–(\frac{x}{2}\left)\right]=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}$

Number of deer drinking water from the pond =9

$x–(\frac{x}{2}+\frac{3x}{8})=9$

$x–\frac{7x}{8}=9$

$\frac{x}{8}=9$

Multiplying both sides by 8,

x=72.

The total number of deer in the herd is 72.

9  A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Solution: Let The grand daughters age be x years then grandfather’s age will be 10x years.

According to the problem,

Grandfather’s age = Granddaughter’s age +54 years

10x=x+54

Transposing x to L.H.S,

10x-x=54

9x=54

x=6.

Granddaughter’s age =x years=6 years.

Grandfather’s age=10x=(10×6) years=60years.

10  Aman’s age is 3 times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Solution:  Let aman’s son age =x years.

Then aman’s age=3x years.

Ten years ago their ages was x-10, 3x-10 years respectively.

According to problem 10 years ago 

Aman’s age=5×aman’s son’s age.

3x-10=5(x-10)

3x-10=5x-50

Transposing 3x to R.H.S and 50 to L.H.S,

50-10=5x-3x

40=2x

x=20

Aman’s son’s age=x years=20 years.

Aman’s age =3x years=(3×20) years=60 years.

 

Exercise 2.5

Solve the following linear equations.

$1 \frac{x}{2}–\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

$Given \frac{x}{2}–\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

Solution: LCM of denominators 2,3,4 and 5 is 60

Multiply both sides by 60.

$60(\frac{x}{2}–\frac{1}{5})=60(\frac{x}{3}+\frac{1}{4})$

⇒30x-12=20x+15

Transposing 20x to L.H.S and 12 to R.H.S

⇒30x-20x=15+12

10x=27 

$x=\frac{27}{10}$

$2 \frac{n}{2}–\frac{3n}{4}+\frac{5n}{6}=21$

Solution: LCM of denominators 2,4 and 6 is 12

Multiply both sides by 12

$12(\frac{n}{2}–\frac{3n}{4}+\frac{5n}{6})=12\times 21$

6n-9n+10n=252

7n=252

$n=\frac{252}{7}$

n=36

$3 x+7–\left(\frac{8x}{3}\right)=\frac{17}{6}–\frac{5x}{2}$

$Given x+7–\left(\frac{8x}{3}\right)=\frac{17}{6}–\frac{5x}{2}$

 LCM of denominators 2,3and 6 is 6

Multiply both sides by 6

$6[x+7–(\frac{8x}{3}\left)\right]=6[\frac{17}{6}–\frac{5x}{2}]$

6x+42-16x=17-15x

-10x+42=17-15x

Transpose 15x to L.H.S and 42 to R.H.S

-10x+15x=17-42

5x=-25

x=-5

$4 \frac{(x–5)}{3}=\frac{(x–3)}{5}$

$Given \frac{(x–5)}{3}=\frac{(x–3)}{5}$

LCM of denominators 3 and 5 is 15

Multiply both sides by 15

$15\times \frac{(x–5)}{3}=\frac{(x–3)}{5}\times 15$

5(x-5)=3(x-3)

5x-25=3x-9

Transposing 3x to L.H.S and 25 to R.H.S

5x-3x=25-9

2x=16

x=8

$5 \frac{(3t–2)}{4}–\frac{(2t+3)}{3}=\frac{2}{3}–t$

S0lution:

$Given \frac{(3t–2)}{4}–\frac{(2t+3)}{3}=\frac{2}{3}–t$

 LCM of denominators 4 and 3 is 12

Multiply both sides by 12

$12[\frac{(3t–2)}{4}–\frac{(2t+3)}{3}]=12[\frac{2}{3}–t]$

9t-6-8t-12=8-12t

t-18=8-12t

12t+t=8+18

13t=26

t=2

$6 m–\frac{(m–1)}{2}=1–\frac{(m–2)}{3}$

Solution:Given 

$m–\frac{(m–1)}{2}=1–\frac{(m–2)}{3}$

LCM of denominators 2 and 3 is 6

Multiply both sides by 6

$6[m–\frac{(m–1)}{2}]=6[1–\frac{(m–2)}{3}]$

6m-3m+3=6-2m+4

3m+3=10-2m

3m+2m=10-3

5m=7

$m=\frac{7}{5}$

Simplify and solve the following linear equations.

7  3(t-3)=5(2t+1)

Solution: 3t-9=10t+5

3t-10t=5+9

-7t=14

t=-2

8 15(y-4)-2(y-9)+5(y+6)=0

Solution: 15y-60-2y+18+5y+30=0

18y-12=0

18y=12

$y=\frac{12}{18}$

$y=\frac{2}{3}$

9  3(5z-7)-2(9z-11)=4(8z-13)-17

Solution: 15z-21-18z+22=32z-52-17

-3z+1=32z-69

69+1=32z+3z

70=35z

Dividing both sides by 35

2 =z

10  0.25(4f-3)=0.05(10f-9)

Solution : 1f-0.75=0.5f-0.45

f-0.5f=0.75-0.45

0.5f=0.3

Dividing both sides by 0.5

f=3 /5=0.6

Exercise 2.6

Solve the following equations.

$1 \frac{(8x–3)}{3x}=2$

Solution: 

$Given \frac{(8x–3)}{3x}=2$

on multiplying both sides by 3x.

$3x\left[\frac{8x–3}{3x}\right]=3x\left(2\right)$

8x-3=6x

8x-6x=3

2x=3

$x=\frac{3}{2}$

$2 \frac{9x}{7–6x}=15$

Solution : 

$Given \frac{9x}{7–6x}=15$

on multiplying both sides by 7-6x

$(7–6x)\left[\frac{9x}{7–6x}\right]=(7–6x)15$

9x=105-90x

9x+90x=105

99x=105

$x=\frac{105}{99}=\frac{35}{33}$

$3 \frac{z}{(z+15)}=\frac{4}{9}$

Solution : 

$Given \frac{z}{(z+15)}=\frac{4}{9}$

On multiplying both sides by 9(z+15)

9z=4(z+15)

9z=4z+60

9z-4z=60

5z=60

On dividing both sides by 5

z=12

$4 \frac{(3y+4)}{(2–6y)}=–\frac{2}{5}$

Solution:

$Given \frac{(3y+4)}{(2–6y)}=–\frac{2}{5}$

On multiplying both sides by 5(2-6y)

5(3y+4)=-2(2-6y)

15y+20=-4+12y

15y-12y=-4-20

3y=-24

On dividing both sides by 3

y=-8

$5 \frac{(7y+4)}{(y+2)}=–\frac{4}{3}$

Solution: 

$Given \frac{(7y+4)}{(y+2)}=–\frac{4}{3}$

On multiplying both sides by 3(y+2)

3(7y+4)=-4(y+2)

21y+12=-4y-8

21y+4y=-8-12

25y=-20

On dividing both sides by 25

$y=–\frac{20}{25}=–\frac{4}{5}$

6  The ages of Hari and Harry are in the ratio 5: 7. Four years from now the ratio of their ages will be 3:4 find their present ages.

Solution: 

Let the common ratio between the ages be x.

Then heri’s age and herry’s age will be 5x years and 7 years respectively.

And four years later, their ages will be (5x+4) and (7x + 4) respectively.

According to the problem,

 

$\frac{5x+4}{7x+4}=\frac{3}{4}$

.

4(5x+4)=3(7x+4)

20x+16=21x+12

20x-21x=12-16

-x=-4

x=4

 Hari’s age =5x =5(4)=20 years

Harry’s age =7x =7(4)=28 years

∴Hari’s age and Harry’s age are 20 years and 28 years respectively.

7  The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3 /2 find the rational number.

Solution : Let the numerator of the rational number be x.

Then its denominator will be x+ 8 the rational number will be x /x+ 8 

According to the problem,

$\frac{x+17}{x+8–1}=\frac{3}{2}$

$\frac{x+17}{x+7}=\frac{3}{2}$

2(x+17)=3(x+7)

2x+34=3x+21

2x-3x=21-34

-x=-13

x=13. 

Numerator of rational number=x=13

Denominator of rational number=x+8=13+8=21

Required rational number=13 /21.

SEMRUSH